Let me direct you to the “Snow Job” logical fallacy. 
Snow job or not, I have to defer. There may be something to what you say.
Do you want the keywords ‘weight distribution’ and ‘inertia’ auto banned as billybobs?
Hahaha! That would be hilarious. 
The question was not this complicated. We understand that you are either very good at physics or very good at googling physics. I do appreciate the information you have provided, but this is yoyo, not a knowledge bowl. keep it simple, keep it fun, keep it on subject. Say what you need and try not to argue, i mentioned that in my initial post.
ok
another #oh teh humanatee# moment when people call serious subjectivity for offense. N00btips at life: neither scientists nor trolls are as easily offended as you think lol. If I did want to offend people, I could have just left ‘‘oh duh look another of these pointless bs topics’’, absolutely no need to type this much.
If I did sound anywhere near arrogant in this thread, I do apologize and I also think its because the people who hold opposing views did not care to backup their opinion with any kind of reasoning, evidence or proof. I said what I need. I presented my view point, person 1 came and said im wrong. I explain my point again, person 2 comes says im wrong. Its just this over and over, and nobody wanted to explain to me why they think im wrong or they r right.
Opinions without reason is bullderp especially on this fundamental a level, which itself wasnt even brought up by me. You even stated in the OP yourself that you want these opinions ‘less subjective’.
And in the same way you yourself didnt even explain why the question is not this complicated and my post is not on subject. If you sound this sure about these opinions of yours, you must have very good reasons behind them. Or it is you yourself that came with a random ad hominem accusation against me lol
/derp
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Hey man. Just because I didn’t bust out equations doesn’t mean I didn’t provide explanations and reasoning… I was just willing to let it drop.
Here’s what I don’t understand about the equation. You’re picking r=x to refer to your P1-3 series of points (which I assume are all on one yoyo?). It doesn’t make sense given that we’re talking about weight distribution, not the radius of a given point. And in the comparison, you assign different r values. If the P is constant, isn’t the radius at that point also constant?
In other words, from two yoyos, you can still get the same r value, which does not correlate to mass at that point… :-/ And in the event that one yoyo has a larger diameter, your mass at any given radius can still be zero… it’s a virtual yoyo at that point, but it doesn’t matter.
If the P series are NOT found on the same yoyo, then you’re talking about tethers of different lengths… which is another argument altogether. With a constant mass at the end of a tether, the shorter the tether the more quickly the mass will complete one rotation around point O.
The real question is that if a ring/band of mass at each of the points P1, P2, P3 have differing masses (ie. the weight distribution is different) but contribute (along with the rest of the object) to the same total mass, with the object accelerate more rapidly or not?
The equations, while impressive, still leave more questions than they give answers. Those of us who decide to forgive your Snow Job fallacy and give it an honest try are left wondering… OK, so you’re calculating what… total energy? Shouldn’t you be calculating velocity at given energy output? I’m not saying I’m right, I’m saying you need to explain what you’re solving for. And if we’re talking about yoyos, at least put a picture of the yoyo so that we can know for sure that your P1, 2, and 3 are ON the yoyo.
P1 and P3 represent the yoyos rims (=thickest/heaviest part where the weight is the most concentrated, but not necessarily outermost), P2 is the center axle where the string is attached, and is really just there for position’s reference (it does not even need to have mass).
When P1=2 and P3=4, the ‘yoyo’ has a radius of 1 unit (weight is most concentrated 1 unit from the center/axle). When P1=1 and P3=5 the yoyo has a radius of 2 and is thus more rim weighted. In both cases the tether length is always P2=3, which could also be any arbitrary number.
Assuming this yoyo is hanged vertically, P1 should be right above the werrd diamond logo, P2 at the center spike, and P3 near the number 30.
Don’t P1 and P3 have the same r, then? And why r? The more important thing is the mass.
As presented (though not necessarily as it exists in your mind) the reasoning is still flawed. More interesting is the axle (let’s call it P2 to keep your convention) and then a P1 where the WERRD part is and a P3 at the rims. But, that only works if we can assume that all intervening masses are identical. If you were to randomly pick out a P4 or any other r value, mass would have to be identical or the entire equation becomes arbitrary and meaningless. Only then, if you swap the P1 and P3, you have changed the weight distribution between two yoyos.
Once you do that, they WILL spin-up differently. Their moment of inertia for rotation will be different.
But once attached to a tether, will the object as a whole accelerate differently around O? I don’t think the equation is proving that it does.
Just to be perfectly clear: at this point in time I’m actually almost convinced. I’m holding out because I prefer clarity over obscurity. And if you’re already on the subject, which do you think is faster? The yoyo with the new P1 having more mass (mid-weighted) or P3 having more mass (rim-weighted)?
I assumed all P’s have the same mass (P2’s can be excluded). P1 and P3 are always symmetrical about P2. Both P1 and P3 represent the rim; they are just opposite to each other across P2.
The red set approximates the real weight distribution of the yoyo in the picture, and the green set represents a hypothetical center weighted version where the heaviest ring is only as big as the inner cup (think of the 3yo3 Ti5)
The entire premise seems to be flawed. The most important thing: the yoyo is spinning. Having P1 and P3 be at the same distance from the axle doesn’t work for the model you’re trying to demonstrate. I understood the P to be “if you drew a radius line, and pick a point, the ring corresponding to this point”, making P1 and P3 equal. If you mean just one specific point, then the entire model is flawed as the yoyo’s spin isn’t accounted for.
Then in the “center-weighted” version, what you’re doing is making the distance from O effectively shorter. The green P3 will be able to make a revolution around O faster than the red P3. I think that’s a given and not anything I’d bother to debate.
Finally, though, choosing radius means selecting a unit of measurement not under discussion. The mass of all the Ps can’t be the same if we’re discussing weight (mass) distribution.
The proposed model just don’t work! Especially given that the final calculation is about Energy rather than acceleration.
IIrc everyone agreed in the end in the other thread that spin has no effect on a round object, so I just assumed this as a general condition and removed spin altogether. Plz provide references from that thread if you has come up with more evidences that suggest the contrary.
Angular speed (ω or rpm) is the controlled variable in this model, not linear speed. R2 (string length) is also kept constant. The green Ps and the red Ps all revolve around O with the same rpm, but their linear speed is proportional to each one’s radius. The green and red P2s do have the same linear speed, but that is only because they have the same R2.
What i m trying to demonstrate is when both R2s revolve around O with the same speed (both angular and linear), the red set contain more energy than the green set. Irl this energy is provided by the driving force and the distance (also time if it is very short) this force drives the object over (W=F*s). Both F and s represent how much you need to move your hand/finger to accelerate the yoyo to such a speed. So the interpretation of the definition of yoyo speed irl is that a faster yoyo require less force input to reach a certain speed than a slower yoyo.
We have ignored all mass inbetween R1 and R3, because the total mass, and hence number of point masses, are equal in both yoyos, and when comparing the green yoyo with the red yoyo, we can always find a pair of corresponding green points for every given pair of red points, such that R1(red) is always less than R1(green), etc. If E1>E2 for every such pair, it must also hold true when summed(integrated) across the entire yoyo. In fact, such an integration gives exactly what we like to call the mass moment of inertia.
Note that even if the yoyo is spinning, it s spinning around P2 so this symmetry is still valid: All points at the same distance from P2 (the collection of which forms a circle/ring across P1 and P3) has the same mass(or concentration of unit mass points). It doesnt matter which exact P1 and P3 are there at one specific moment. They all weigh the same and have the same distances R.
Too much math.
Lets just dumb it out a little with a different question?
What do you prefer, floaty, solid, or fast?
The way I see it there are two distinct dynamic forces at work. the first is a center of mass rotating about a central point at a distance R. The other is a mass rotating on an axis.
Clearly the mass of the object spinning is less important than the moment of inertia, or the distribution of mass. This is illustrated by a spinning figure skater. If he/she holds her arms outward, they spin slowly. If they pull them in, they spin faster. The skater has not changed their mass, only the way in which it is distributed.
The speed of this axial spin is important because it causes angular momentum. Angular momentum is a force that acts against any external system forces and creates a resistant force to any changes in speed or direction. This is why a Frisbee flies through the air with stability.This interaction is dynamic, not linear. That is why it is resisting easy explanation.
You cannot disregard the angular momentum of the yoyo. Its force will be contrary to the force of gravity as it spins in a circle (unless you are doing horizontal). This, coupled with the rotational speed due distribution of mass, will result in multiple forces acting on the yoyo, at any given time as it moves; defying easy explanation.
It just shows what an amazing calculating machine that the universe is. Its all done in real-time too. 
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Angular momentum is NOT a force! Please go reread your physics textbook.
db
There HAS to be an effect, or P1 and P3 are in all ways equal and it therefore becomes irrelevant to distinguish between them. In your model, P1 and P3 exchange positions about 100-150 times per second and all points in-between during the intervening time.
I do think you’re on the right track to enlightening me, I just don’t think you’ve chosen the right model to explain it. I don’t mean that in the snotty way that it sounds, it’s just that the P1/3 thing and the choice of “r” as a unit for explaining the effect of weight distribution is throwing me right off.
What is starting to make sense in my mind is that if you have a theoretical “ring” of mass on the outside of an otherwise uniformly-distributed yoyo, vs. a ring of mass more towards the axle, the play characteristics are obviously different, but the speed characteristics might be, too. A higher concentration of mass is more towards the “O”, which… maybe…? makes it play faster? That’s the unanswered question for me. Does it?
The tricky part is this:
While pushing weight to the outside puts more weight further from “O”, it simultaneously moves the same mass more TOWARDS the “O”. And with a center-weighted yoyo, you have mass not as far from “O” on the outside (theoretically faster, if it was not a cocentrically spinning object), but then on the inside it is also further away from “O” than the rim-weighted model.
Is it not just a wash?
The whole “r” thing is starting to imply two uniformly-distributed discs with different outside diameters… which is not what was under discussion and I think becomes a given.
Anecdotally, it’s worth noting that the length from O to P2 (R2? where are we now? The axle!) is not actually the same when the yoyo changes size. Most yoyoers tend to cut their strings so that the total length from outside edge to the loop is equivalent.
I wrote up some basic definitions that describe how I use the terms. Keep in mind that a yo-yo does not have to be described with only one of these words. This is what I mean when I use the terms on the board.
Float - I call this yo-yo “hang time.” It describes how long the yo-yo appears to hover in the air during play. A matter of perception, and subjective. (example, Model 10)
Solid - For me, solid is how heavy the yo-yo feels when it hits the string. It will give the feeling of playing with a more dense yo-yo. A matter of perception, and subjective. (example, Genesis)
Fast - Speed is a formula to me, which can be determined, in part, by the speed of the player, and the length of the string, among other things (as posted by Gambit earlier). But, all things being equal, ask yourself “how does the yo-yo play compared to other yo-yos?” If you play with the same length of string, and do your tricks at the same speed, does one yo-yo seem faster than the average? If so, that yo-yo might be referred to as “fast.” Also, a matter of perception and subjective. (example, Volume)
A yo-yo might have elements of floaty and solid, if it seems to hover in the air a bit, but still hits the string with a nice presence too. Not easy to get that blend, but it has been achieved. I believe the Cypher is a hybrid of float and solid. More floaty than solid though.
Also, when many players have played the same yo-yo, you might notice that over time, there is a general consensus about whether the yo-yo is fast, solid, or floaty. That consensus allows us to share how the yo-yo feels with others, but I would still not consider these perceptions a fact. But, if there is a general consensus, chances are the yo-yo feels that way to “most” people. That is why these terms are very useful in reviews.
^ That is how I would explain those terms.
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You dont need to fix P1 and P3 to any real point on the yoyo. At any given point in time, you can find a P1 and a corresponding P3, and their mass are always equal. Thats enough.
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This is where the squared term in the formula comes into play. While it is true that R2-R1=R3-R2, also R2^2 - R1^2 <R3^2 - R2^2. As such, the momentum-increasing effect of P3’s outward shift always surpasses the momentum-decreasing effect of P1’s shift toward O.
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I thought about this writing my previous post but ultimately deemed it too trivial and arbitrary. Do note, however, that even if people do this, they are shifting the masses even further away from O with a larger yoyo, and this will slow the yoyo further down instead of speedign it up.
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It was very hard to tell until now that we agree that P1 and P3 are equal points.
That might be on me. -
momentum and acceleration are not the same thing. You can accelerate something very rapidly but it may not have momentum. Acceleration is the important thing to talk about here.
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Hey, you’re the one who mentioned it.
But no, the point is that if people cut the tether from the outside edge to their ideal length, P3 is always at the same distance from O.
Are you getting tired of this?
How about quitting the hair-splitting? I wrote:
“Angular momentum is a force that acts against any external system forces”
Since Force is something that describes the interaction of two bodies, I am speaking about the Forces between the string, the rotational body and gravity. These are three interactions that will result in forces. The rotational momentum is a measure of the FORCE vectors affecting it at any time. The resulting equation is of course, not a FORCE. it describes the rotational behavior of a set of particles. That behavior is determined by the FORCES at work on these particles at any given time
So lets focus in on the generic use of the word: force. Angular momentum is the reason that a Frisbee stays stable in flight. The angular momentum FORCES the Frisbee to stay stable by resisting the FORCE of air pressure against it. Notice the use of the English term FORCE and FORCES. It does not denote a physics term, it is a word that can be an adverb or a verb. It describes.
Kind of like “Jerk”. It can be an action, a description of an action; … or a person.